Examples of reflective projective billiards and outer ghost billiards
aa r X i v : . [ m a t h . D S ] A p r Examples of reflective projective billiards and outer ghost billiards
Corentin FierobeApril 14, 2020
Abstract
In the class of projective billiards, which contains the usual billiards, we exhibit counter-examples toIvrii’s conjecture, which states that in any planar billiard with smooth boundary the set of periodic orbitshas zero measure. The counter-examples are polygons admitting a 2-parameters family of n -periodicorbits, with n being either 3 or any even integer grower than 4. Contents -reflectivity of the right-spherical billiard 43 Two -reflective projective billiards 5 In the theory of billiard, describing properties of the periodic orbits of the billiard inside a domain is thesubject of many studies. A famous conjecture, due to Ivrii states the following: in any smooth euclideanplanar billiard, the set of periodic orbits has zero measure . The proof of the conjecture was made in the caseof a billiard with a regular analytic convex boundary, see [14]. The conjecture is also true for 3-periodicorbits, and was proved in [1, 9, 10, 15, 16]. For 4-periodic orbits, it was proven in [7, 8] and a completeclassification of 4-reflective (defined later) complex analytic billiards was presented in [6]. Ivrii’s conjecturewas also studied in manifolds of constant curvature: it was proven to be true for k = 3 in the hyperbolicplane H , see [3]; the case of the sphere S was apparently firstly studied in [2], as quoted in [3] but we werenot able to find the correponding paper. The sphere is in fact an example of space were Ivrii’s conjecture isnot true, and we can find a classification of all counter-examples in [3].b γ γ O Figure 1: Left: a curve b endowed with a field of transverse line. Center: a convex closed curve γ with a fieldof transverse lines. Right: the same curve γ with a so-called centrally-projective field of transverse lines.1n this paper we study a generalization of usual billiards, as it is described in [11]. We consider the socalled projective billiards , which are billiards having their boundaries endowed with a field of transverse lines,called projective field of lines (see Figure 1), and defining at each point a new reflection law which we call projective reflection law : suppose that at a point p of a curve γ , the transverse line at p is L . A line ℓ hitting γ at p will be reflected into a line ℓ ′ if the lines ( ℓ, ℓ ′ , L, T p γ ) are harmonic, meaning that their cross-ratio is −
1. We recall that the cross-ratio of four lines through the same point p is defined as the cross-ratio of theirintersection with a fifth line not passing through p , and this doesn’t depend on the line taken. Note that thisdefinition makes the projective plane RP a convenient space to study projective billiards.We can therefore study the billiard dynamics on projective billiard chosing this new law to determine theway a particle bouces inside a table. Note that, given any closed curve γ , the usual billiard dynamic inside γ is also a projective one by choosing the transverse lines to be the normal lines to γ . In this paper we studyexamples of projective billiards inside polygons. They are defined as follows:EXAMPLE 1. Consider a smooth convex planar curve γ and an other point O inside γ called the origin.Define a transverse field of lines on γ by taking the lines passing through the origin, see Figure 2. As in[11], this billiard is called centrally-projective and has many interesting properties, including the fact thatthe billiard transformation has an invariant area form ([11], Theorem D).In the case when γ is a polygon P with n -vertices, P , . . . , P n − , we call this billiard a centrally-projectivepolygon , and denote it by CPol( O ; P , . . . , P n − ). When the poygon P is regular and the origin is theintersection of its great diagonals, we call it centrally-projective regular polygon . γ O P P P P OL L L L Figure 2: Left: a convex closed curve γ with a so-called centrally-projective field of transverse lines throughan origin O . Right: a centrally-projective quadrilateral P P P P with origin O .EXAMPLE 2. Consider a triangle P P P in the (projective) plane. On a side P i P i +1 , define a field oftransverse lines by taking the lines passing through the remaining vertex of the triangle, P i − , see Figure 3.We call this projective billiard right-spherical billiard based at ( P , P , P ) . and denote it by S ( P , P , P ). P P P L L L Figure 3: The right-spherical billiard S ( P , P , P ) with each one of its fields of transverse lines2ts name comes from a famous construction on the two-dimensionnal sphere S : consider a triangle Q Q Q on S having right angles at all its vertices (its area is one eighth of the total surface of S ). Thistriangle is a well-known example of 2-reflective billiard on the sphere, as explained in [2, 3]. Now consider theprojection of S from its center to a plane P tangent to the sphere at a point located inside Q Q Q : Q Q Q is projected on a triangle P P P of P , and the normal geodesics to the sides of Q Q Q are projected intothe projective field of lines making P P P a right-spherical billiard.Here we consider a simplified version of the billiard dynamic, where the name ghost come from, and whichis enough to study examples of curves having the reflectivity property (defined later). In the particular caseof a polygon P = P · · · P n − endowed with a field of transverse lines on each of its sides, we extend thedefinition of edges of P to be the whole lines supporting the usual edges. We force the successive bounces ofa billiard trajectory to be on successive edges of P , by forgetting any possible obstacles on the ball trajectorybetween two consecutive edges. More precisely, fix two distinct points M ∈ P P and M ∈ P P on twoconsecutive edges. We define the orbit of ( M , M ) as follows: Definition 1.1.
The orbit of ( M , M ) is the sequence ( M k ) k ∈ Z where for each k , M k is a point on P k P k +1 (where k is taken modulo n ) and the lines M k − M k and M k M k +1 are symmetric with respect to the projectivereflection law at M k .Let m = kn be a positive mutliplier of n . The orbit is called m -periodic if M m M m +1 = M M . Thecorresponding projective billiard is said to be m -reflective if there is a non-empty open subset U × V ⊂ P P × P P of ( M , M ) whose orbits are m -periodic.In the following, we use these examples to construct reflective projective billiards (having the additionalparticular property that U × V = P P × P P , which can be seen as a consequence of analyticity of thebilliard map). First, since the right-spherical billiard is obtained by projecting a certain 3-reflective billiardof S on the plane, we expect that it is also 2-reflective (see Figure 4). Indeed, we have the Proposition 1.2.
The right-spherical billiard, S ( P , P , P ) , is -reflective. P P P M M M L L L Figure 4: The spherical billiard S ( P , P , P ) and a triangular orbit ( M , M , M ) obtained by reflecting anysegment M M two timesIn Section 3, we construct classes of 4-reflective projective billiards which cannot be deduced from oneanother. More precisely, we build the first one by gluing two right-spherical billiards. The second one is acentrally-projective polygon, as stated by Proposition 1.3.
Let P , P , P , P be a non-degenerate quadrilateral and O the intersection point of itsdiagonals. The centrally-projective polygon CPol ( O ; P , P , P , P ) is -reflective. In Section 4, we generalize the case n = 4 to any even integer n ≥
4, with restrictions on the polygonhowever:
Proposition 1.4.
Let n ≥ be even. Let P · · · P n − be a regular polygon and O be the intersection point ofits diagonals. The centrally-projective regular billiard CPol ( O ; P , . . . , P n − ) is n -reflective. n even. However, if one allows two times more bounces, it becomes true, andnot only for regular polygons. More precisely, we state the following Proposition 1.5.
Let n ≥ be odd. Let P · · · P n − be a polygon and O be any point not lying on the edgesof P . The centrally-projective billiard CPol ( O ; P , . . . , P n − ) is n -reflective. The results are proved in the following order: we prove Proposition 1.2 in Section 2, Proposition 1.3in Section 3 (and present another example of 4-reflective billiard derived from the right-spherical model),Proposition 1.4 in Section 4, and Proposition 1.5 in Section 5.Knowing these examples, many questions arise: are there polygonal examples of n -reflective projectivebilliards with n odd? Can we find centrally-projective billiards which are n -reflective but not built onpolygons? Another more difficult question would be: what are the n -reflective projective billiards (with therequirement that the boundary has a certain class of smoothness)? Answering this question would be a greatstep in solving Ivrii’s conjecture. In [4], we show that the only 3-reflective projective billiard table withanalytic boundary is the right-spherical billiard, as described in example described in Section 2.We would like to conclude this introduction by saying that you can find, on the webpage of the author(which was [5] at the time of writing), a simulation of projective billiards in polygons in which you canvisualize their dynamics. -reflectivity of the right-spherical billiard In this section, we prove Proposition 1.2: the right-spherical billiard is 3-reflective. We first introduce somenotations:Let P , P , P be three points not on the same line. For i = 0 , ,
2, let γ i be the line P i P i +1 . For any M ∈ γ i , let L i ( M ) be the line M P i +2 ( i is seen modulo 3), that is L i ( M ) is line joining M and the onlypoint P j which do lie on γ i . The projective billiard table defined by the γ i and the L i is the right-sphericalbilliard S ( P , P , P ) (see Figure 3). Proposition 2.1.
Any ( M , M ) ∈ γ × γ with M = M determines a -periodic orbit inside the right-spherical billiard S ( P , P , P ) .Proof. This proof was found by Simon Allais in a talk we had about harmonicity in a projective space. Let M ∈ γ be such that M M , M M , γ , L ( M ) are harmonic lines. Define M ′ ∈ γ similarly: M M , M M ′ , γ , L ( M ) are harmonic lines. P P P M M M L L AM ′ Figure 5: As in the proof of Proposition 2.1, both quadruples of points ( M , A, P , P ) and ( M ′ , A, P , P )are harmonic, hence necessarily M = M ′ . 4et us first show that necessarily M = M ′ (see Figure 5). Consider the line γ and let A be its pointof intersection with M M . Let us consider harmonic quadruples of points on γ . By harmonicity of theprevious defined lines passing through M , the quadruple of points ( A, M , P , P ) is harmonic. Doing thesame with the lines passing through M , the quadruple of points ( A, M ′ , P , P ) is harmonic. Hence M = M ′ since the projective transformation defining the cross-ratio is one to one.Now let us prove that the lines M M , M M , γ , L ( M ) are harmonic lines. Consider the line γ : M M intersects it at a certain point denoted by B , M M at M , γ at P and L ( M ) at P . But thequadruple of points ( B, M , P , P ) is harmonic since there is a reflection law at M whose lines intersect γ exactly in those points. -reflective projective billiards In this section we construct two classes of 4-reflective projective billiards: the first one is obtained by gluingtogether two right-spherical billiards, the second one is obtained from a centrally-projective quadrilateralwhose origin is the intersection point of its diagonals. γ ′ γP P M M M ′ M ′ M L ′ L Figure 6: billiard deduced from the right-spherical billiard S ( P , P , P ), where the point P is at infinity,hence the lines P P , P P and all tranverse lines L ( M ) on P P are parallel, making the reflection law on P P as the usual one.In R , consider two distinct parallel lines γ and γ ′ . Choose a normal line intersecting γ at a point P and γ ′ at a point P . On γ , define a field of transverse lines by taking the lines L ( M ) going from a point M ∈ γ to P . Do the same on γ ′ with P to form the field of lines L ′ . Consider the projective billiard dynamicsmade by a particle bouncing alternatively γ and γ ′ with their respective fields of transverse lines. Call thisconstruction the converging mirrors . We claim that Proposition 3.1.
The converging mirrors are -reflective.Proof. Complete R with a line at infinity ℓ to form the projective plane. Consider the point P ∈ ℓ where γ and γ ′ intersect. Choose any orbit ( M , M , M ) of the spherical billiard S ( P , P , P ), where M ∈ γ , M ∈ P P and M ∈ γ ′ (see Figure 6). Now consider the axial reflection with respect to the line P P .It leaves P P , γ , γ ′ and P invariant, hence the orbit ( M , M , M ) is transformed into another orbit( M ′ , M , M ′ ) of S ( P , P , P ), where M ′ ∈ γ and M ′ ∈ γ ′ .But since the transverse lines on the side P P of S ( P , P , P ) are orthogonal to the line P P (they passthrough P , as well as γ and γ ′ ), the reflection law on P P is the usual one: the lines M M and M M make the same angle with P P . Hence M , M , M ′ are on the same line, and so are M , M , M ′ . Therefore, M , M ′ , M ′ , M is a 4-periodic orbit. In this subsection, we prove Proposition 1.3. We first introduce some notations:5et P , P , P , P be four points, no three of them being on the same line. For i = 0 . . . γ i be the line P i P i +1 . Write O to be the point of intersection of the lines P P and P P (diagonals).For any M ∈ γ i , let L i ( M ) be the line OM . The projective billiard table defined by the γ i and the L i iswhat we call the centrally-projective quadrilateral CPol( O ; P , P , P , P ) (see Figure 7). P P P P OL L L L Figure 7: The centrally-projective quadrilateral CPol( O ; P , P , P , P ) with each one of its fields of transverselines Proposition 3.2.
Any ( M , M ) ∈ γ × γ with M = M determines a -periodic orbit of the centrally-projective quadrilateral CPol ( O ; P , P , P , P ) .Proof. Let M ∈ γ such that M M is reflected into M M by the reflection law at M . Let M ∈ γ suchthat M M is reflected into M M by the reflection law at M . Let M ′ ∈ γ such that M M is reflected into P P P P OM M M A B M = M ′ C ′ D ′ A ′ B ′ L L L L Figure 8: The centrally-projective quadrilateral CPol( O ; P , P , P , P ) with a periodic orbit obtained byreflecting M M three times. Here the notations are the same as in the proof of Proposition 3.2.6 M ′ by the reflection law at M . Denote by d the line reflected from M M by the projective reflectionlaw at M . We have to show that d = M M ′ .First, let us introduce a few notations (see Figure 8). Consider the line M M ; it intersects: the line P P at a point B and the line P P at a point A ′ . Now consider the line M M ; it intersects: the line P P at apoint B ′ and the line P P at a point A . Finally let C ′ be the intersection point of M M with P P and D ′ the intersection point of M M ′ with P P .Then, notice that by the projective law of reflection at M , the quadruple of points ( A, A ′ , P , O ) isharmonic. Since the points P , A ′ , O correpond to the lines γ , M M , L ( M ), the previously definedreflected line M M ′ needs to pass through A in order to form a harmonic quadruple of lines. The sameremark on the other diagonal leads to note that M M passes through B .Now by the reflection law at M , one observe that the quadruple of points ( A, C ′ , O, P ) is harmonic.But γ passes through P , M M through C ′ and L ( M ) through O . Hence d needs to pass through A .Then, by the reflection law at M , one observe that the quadruple of points ( B, D ′ , O, P ) is harmonic. But γ passes through P , M M through B and L ( M ) through O . Hence d needs to pass through D ′ .Therefore we conclude that d = AD ′ = M M ′ . Remark . Notice that the spherical billiard cannot be deduced from a usual billiard on the sphere by thesame construction as described for the right-spherical billiard.
In the following, we prove Proposition 1.4. We first introduce some notations:Let n = 2 k , k ≥
2, an even integer. Let P be an n -sided regular polygon of radius 1, whose vertices areclockwise denoted by P , P , . . . , P n − . In the following, each time we refer to the index i , we will consider itmodulo n , that is we identify i and its rest when divided by n . Define O to be the intersection of the greatdiagonals of the polygon, which are the lines P i P i + k , i = 0 , . . . , k −
1. Construct a field of transverse lines L i on γ i := P i P i +1 by setting L i ( M ) = M O (the line joining the basepoint M and O ). The projective billiard tabledefined by the γ i and the L i is what we call the centrally-projective regular polygon CPol( O ; P , . . . , P n − ),see Figure 9.Let ( M , M ) ∈ P P × P P with M = M . We want to show that the orbit ( M m ) m ∈ Z of ( M , M ) is n -periodic. We first prove the P P P P P P OM M M M M M L L L L L L Figure 9: A centrally-projective polygon CPol( O ; P , . . . , P ) and a piece of trajectory after four projectivebounces. Lemma 4.1.
Fix an m ∈ N and consider the great diagonal L m = P m P m + k . Then for any r ∈ N , the lines M m − r − M m − r − and M m + r M m + r +1 intersect L m at a same point. m P m − r P m − r − P m − r − P m + r P m + r +1 P m + r +2 M m − r − M m − r M m − r − M m + r − M m + r +1 M m + r L m Figure 10: As in the proof of Proposition 4.2, since the lines M m − r − M m − r and M m + r − M m + r intersect L m at the same point, the lines M m − r − M m − r − and M m + r M m + r +1 also intersect L m at a same point. Proof.
Let us prove Lemma 4.1 by induction on r . Case when r = 0 : Fix any m ∈ Z . Let A be the intersection point of M m − M m − with L m , A ′ of M m M m +1 with L m and B of M m − M m . Consider harmonic quadruples of points on L m : ( A, B, P m , O )is harmonic by the reflection law in M m − , and ( A ′ , B, P m , O ) is harmonic by the reflection law in M m +1 .Hence A = A ′ which concludes the proof for r = 0. Inductive step: suppose Lemma 4.1 is true for any r ′ < r and let us prove it for r . See Figure 10 fora detailled drawing of the situation. Fix an m ∈ Z . By assumption, we know that M m − r − M m − r and M m + r − M m + r intersect L m at a same point A . Furthermore by symmetry of the regular polygon through L m , the lines P m − r − P m − r and P m + r P m + r +1 intersect L m at the same point. Now, the three following linesthrough M m − r − , M m − r − M m − r , P m − r − P m − r , M m − r − O intersect L m through the same points than thethree following lines through M m + r , M m + r − M m + r , P m + r P m + r +1 , M m + r O . Hence in order to sastisfy theprojection law at M m − r − and M m + r respectively, the lines M m − r − M m − r − and M m + r M m + r +1 shouldintersect at the same point. Hence the inductive step is over and this conclude the proof. Proposition 4.2.
Any ( M , M ) ∈ P P × P P with M = M determines an n -periodic orbit of thecentrally-projective regular polygon CPol ( O ; P , . . . , P n − ) .Proof. We have to show that M − M = M n − M n . We will use Lemma 4.1. First, by setting m = k and r = k −
1, we conclude that the lines M M − and the lines M n − M n intersect L k at a same point denoted by A . Then, by setting m = k + 1 and r = k − M M and M n − M n intersect L k +1 = L at a same point B . Now it is also true that M M − intersect L at B , by Lemma 4.1 and setting m = 1 and r = 0. Hence we have shown that M n − M n = AB = M M − which concludes the proof. Remark . We note that the assumptions that P is regular and that O is on the intersection of the greatdiagonals is fundamental to guarantee the reflectivity. On a simulation, see for example [5], the orbits aredestroyed after small perturbations of P ’s vertices or of the origin. However, the next example we describeis much more stable in this sense. In this section we prove Proposition 1.5. We first introduce some notations:Let n = 2 k + 1, k ≥
1, be an odd integer. Let P be an n -sided polygon, whose vertices are clockwisedenoted by P , P , . . . , P n − . We suppose that any three consececutive vertices are not on the same line.Choose another point O , which do not lie on any line of the type γ i := P i P i +1 ( i is taken modulo n ).Construct a field of transverse lines L i on γ i by setting L i ( M ) = M O (the line joining the basepoint M and O ). The projective billiard table defined by the γ i and the L i is what we call the centrally-projective polygon CPol( O ; P , . . . , P n − ) (see Figure 11). 8e will prove the Proposition 5.1.
Let n = 2 k +1 ≥ . Any ( M , M ) ∈ P P × P P with M = M determines a n -periodicorbit of the centrally-projective polygon CPol ( O ; P , . . . , P n − ) . P P P M M M O M M M Figure 11: A 6-periodic orbit ( M k ) k on a centrally-projective triangle P P P with origin O . The dottedlines are representatives of the transverse fields of lines on the sides of the triangle. Outer ghost billiard associated to a centrally pojective billiard
We can associate to any projective billiard a billiard called dual , or outer billiard , and this construction iswell-explained in [11]. We will explain from the beginning the constructions which we will use, but for now,the already enlightened reader should keep in mind that the version of outer billiard we will use is verysimplified, and the latter should be better called outer ghost billiard . Usual outer billiards are much morecomplicated, see [12, 13] for more details. Let us first recall some notions about polar duality.Fix a point O ∈ R and complete R with a line at infinity λ to form RP . Consider the polar dualitywith respect to the point O . It sends any point p ∈ RP to a line p ∗ and any line ℓ ⊂ RP to a point ℓ ∗ in a bijective way, such that O ∗ = λ and λ ∗ = O . It is involutive in the sense that if you apply the polarduality with respect to O two times, you get the identity: p ∗∗ = p and ℓ ∗∗ = ℓ . Finaly polar duality has theincidence property: p ∈ ℓ if and only if ℓ ∗ ∈ p ∗ .Now as in Figure 12, suppose that we are given four lines of R , ℓ , ℓ ′ , T and L such that– L passes through O ;– The four lines ℓ, ℓ ′ , L, T pass through the same point p = O and are harmonic.By the incidence property, the points ℓ ∗ , ℓ ′ ∗ , L ∗ and T ∗ belong to the line p ∗ , and are harmonic since thecorresponding four lines are. But also L ∗ ∈ O ∗ = λ is at infinity. By harmonicity, both vectors −−→ T ∗ ℓ ∗ and −−−→ T ∗ ℓ ′ ∗ are opposite.We can apply this to a centrally-projective polygon CPol( O ; P , . . . , P n − ): denote by Q , . . . , Q n − thedual points of its sides, where Q k = P k P ∗ k +1 ( k is tajen modulo n ). Choose a projective orbit ( M k ) k ∈ Z ofCPol( O ; P , . . . , P n − ), an consider, for all k , the dual point N k of the line M k − M k ; we call ( N k ) k the dualorbit of ( M k ) k . Now we are ready to define the outer ghost billiard on Q · · · Q n − (see Figure 13): Definition 5.2.
To any point N distinct from Q , . . . , Q n − , we associate the sequence of points ( N k ) k ,called ghost outer orbit of the polygon Q · · · Q n − , and uniquely determined by the relation −−−→ N k Q k = −−−−−→ Q k N k +1 for all k ∈ Z (where here Q k is used for Q k mod n ). The orbit is said to be m -periodic if N m = N .9 LT pℓ ℓ ′ ℓ ∗ ℓ ′∗ T ∗ Figure 12: On the left: four harmonic lines ℓ , ℓ ′ , L and T through p . On the right: their polar duals withrespect to O , T ∗ is in the middle of the segment joining ℓ ∗ and ℓ ′ ∗ , L ∗ is at infinity. Q Q Q N N N N N Figure 13: Five successive points, N k with k = 0 . . .
4, of an outer ghost orbit of Q Q Q .By previous discussion, we have the Proposition 5.3. If ( M k ) k is an orbit of CPol ( O ; P , . . . , P n − ) , its dual orbit ( N k ) k is an outer ghost orbitof Q · · · Q n − . Furthermore, ( M k ) k is m -periodic if and only if ( N k ) k is m -periodic. Therefore, to prove Proposition 5.1, we only need to show the
Proposition 5.4.
Let n = 2 k + 1 ≥ . Any outer ghost orbit of Q · · · Q n − is n -periodic.Proof. Let ( N k ) k be an outer ghost orbit of Q · · · Q n − . We first show the Lemma 5.5.
For k ∈ Z , we have −−−−−→ N k N k +2 = 2 −−−−−→ Q k Q k +1 .Proof. Here we are in a Thales configuration and the proof follows. Indeed, by Chasles relation and theequalities −−−→ N k Q k = −−−−−→ Q k N k +1 , −−−−−−−→ N k +1 Q k +1 = −−−−−−−→ Q k +1 N k +2 , we can write that −−−−−→ N k N k +2 = −−−→ N k Q k + −−−−−→ Q k N k +1 + −−−−−−−→ N k +1 Q k +1 + −−−−−−−→ Q k +1 N k +2 = 2 −−−−−→ Q k N k +1 + 2 −−−−−−−→ N k +1 Q k +1 = 2 −−−−−→ Q k Q k +1 . To conclude the proof, we need to show that N = N n . Indeed by Lemma 5.5 we have −−−−→ N N n = n − X j =0 −−−−−−→ N j N j +2 = 2 n − X j =0 −−−−−−→ Q j Q j +1 k , where n − k , we get n − X j =0 −−−−−−→ Q j Q j +1 = k X j =0 −−−−−−→ Q j Q j +1 + n − X j = k +1 −−−−−−→ Q j Q j +1 and the latter sum can be rewritten with the change of index j ′ = j − k since −−−−−−→ Q j Q j +1 = −−−−−−−−−−→ Q j − n Q j +1 − n = −−−−−−−→ Q j ′ − Q j ′ , to get n − X j =0 −−−−−−→ Q j Q j +1 = k X j =0 −−−−−−→ Q j Q j +1 + k +1 X j ′ =1 −−−−−−−→ Q j ′ − Q j ′ = 0 . which concludes the proof. Proof of Proposition 5.1.
Since any outer ghost orbit of Q · · · Q n − is 2 n -periodic (Proposition 5.4), so isany projective orbit of CPol( O ; P , . . . , P n − ) by Proposition 5.3. Acknowledgments
I’m very grateful to Simon Allais for his explanations about harmonic lines and how to prove Proposition 2.1with this method, and Alexey Glutsyuk for its very helpful advices.
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